1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Conservation of Energy

  1. Apr 4, 2007 #1
    2. Relevant equations

    Total Energy = Kinetic Energy + Potential Energy
    T.E = 1/2mv^2 + m(g)(h)
    3. The attempt at a solution

    T.E at top
    = 1/2mv^2 + m(9.8)(h)

    T.E at bottom
    1/2mv^2 + m(9.8)(h) = 1/2mv^2 + m(9.8)(h/2)

    Please help I have no clue what to do
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2
  4. Apr 4, 2007 #3
    Oh yeah, I had that on my paper, but forgot to type it in.
  5. Apr 4, 2007 #4


    User Avatar
    Homework Helper

    You look to be on the right track. You just need to solve for v2. Are you having trouble with the algebra? Put "g" back in for the 9.8, to make it clearer.
    You can also cancel out m.
    Last edited: Apr 4, 2007
  6. Apr 4, 2007 #5
    1/2mv^2 + m(g)(h) = 1/2m[v]^2 + m(g)(h/2)
    v^2 + (g)(h) - (g)(h/2) = [v]^2

    Is this right? =/ I really suck at simplifying
  7. Apr 4, 2007 #6


    User Avatar
    Homework Helper

    The first line is good. You can simply your gh terms. If you ignore the gh for a moment, you will see that it is really 1-1/2, which is 1/2.

    You have also dropped the 1/2 from the kinetic energy terms again.
  8. Apr 4, 2007 #7
    1/2mv^2 + 1/2 = 1/2m[v]^2
    v^2 + 1 = [v]^2 *multiply both sides by 2 to eliminate 1/2?
    v + 1 = [v]
  9. Apr 4, 2007 #8
    well the help so far has been on the right track.

    you can work this out if you choose, but any increase in energy will come from the additional drop of 1/2y*mg

    so 1/2mv'^2=1/2mv^2 +1/2 mgh,
    so factoring, gives v'=sqrt(v^2+gh)
    may be wrong but don't think you have enough information to solve for h.
  10. Apr 4, 2007 #9
    Okay, I gotta ask, Bambi's carcass??!
  11. Apr 4, 2007 #10


    User Avatar
    Homework Helper

    Oh sorry I should have been clearer...you still need the gh in there. I was just pointing out that gh - (1/2)gh = (1/2)gh. So put a gh in where your 1 is.

    Also v must remain as v^2, since you have

    v^2 + gh = [v]^2

    when you take the square root of both sides, you get

    (v^2 + gh)^(1/2) = [v]

    You can't take the square root of individual terms like you did. You must take it of the entire left side.
  12. Apr 4, 2007 #11
    Robb, I had to laugh as well, I think there is now an industry to make physics problems more entertaining to youth of america. so they dress up the same old dreary problems in ever increasingly bizarre clothes.
  13. Apr 4, 2007 #12


    User Avatar
    Homework Helper

    lol I didn't want to know. :yuck:
  14. Apr 4, 2007 #13
    yess thank you I did it again and got this answer, thank you both so much for all your help.
    Last edited: Apr 4, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook